That the converse is also true, is affirmed by a (more
general) result of Jawhari, Misane and Pouzet [1]}, Proposition IV-4.1. This
characterization is, however, not yet completely satisfactory because
of its partially extrinsic nature: how would we check that a
distance is closed other than by computing the lower cone, which
consists of the words coding potential zigzag factors? Fortunately,
there is an intrinsic way to verify closedness: assume \(y+z\)  and
\(y-z\)  are any two words in a distance \(d(a,b)\)  with common circumfix
\(y \ldots z\)  but different infix letters; then the common subword
\(yz\)  must belong to \(d(a,b)\)  whenever this distance is closed. We
refer to this checking procedure as the "cancellation rule". To give
an example, assume that a closed upper set \(Z\)  of words contains \(+++\) 
and \(---\) . Then, as \(Z\)  is an upper set, we have both \(+-+-+\) 
and \(--+-+\)  in \(Z\) , whence \(-+-+\)  belongs to \(Z\)  by the cancellation
rule. Further, as \(++-+\)  is in \(Z\) , the rule applied to the latter
two words returns \(+-+\) , and since also \(+++\)  is in \(Z\) , so must be
the common subword \(++\) . Interchanging the roles of \(+\)  and \(-\)  we
infer that \(Z\)  contains \(--\) . Repeating essentially the same argument
for the words \(++\)  and \(--\)  proves that \(+\)  and \(-\)  are words in \(Z\) ,
whence \(Z\)  contains the empty word and thus is all of \(\lbrace +,-\rbrace ^{\ast }.\)
